Question

Is the sum from 1 to infinity of n/(n+2) convergent or divergent?

Answer 1

The answer is convergent, but why?

Answer 2

does the limit of the sequence goes to zero ?

Answer 3

No, it goes to 1. Yet the answer on the test is convergent.

Answer 4

its not convergent

Answer 5

\[\frac{n}{n+2}\]
\[\frac{n+2-2}{n+2}\]
\[\frac{n+2}{n+2}-\frac{2}{n+2}\]
\[1-\cancel{\frac{2}{n+2}}^{~0}=1\]

Answer 6

if the sequence does not limit to zero, the series is not convergent

Answer 7

lol need not perform that much algebraic manipulation to see it diverges, clearly the terms are roughly one, so the sum grows linearly, and clearly diverges

Answer 8

lol, but i like to practice archaic devices

Answer 9

1-2/3 + 1 - 2/4 + 1 - 2/5 + 1 - 2/6 + ....

your always adding 1 and subtracting less than 1 ... subtracting even smaller amounts less than 1

its not going to converge

Answer 10

ahh, a sequence is not a series

Answer 11

I've attached the problem, so I'm with you guys, it is divergent, but did I miss anything?

Answer 12

the sequence converges to 1

Answer 13

if the limit of the sequence is finite, then the sequence itself converges

if the limit of the sequence is not 0, then the series (sum of the sequence) diverges

Answer 14

\[\lim_{n \rightarrow \infty} \frac{ n }{ n+2 }=1\] so diverges

Answer 15

I forget, what is the distinction between a sequence and a series?

Answer 16

a sequence is a list of numbers

a series as the sum of that list

Answer 17

if sum is conv. then limit must be 0.... if limit is different than 0 then it is div..

Answer 18

1,2,3,4,5,... is a sequence of numbers; a line up

1 + 2 + 3 + 4 + 5 + .... is a series (a sum of the numbers in the list)

Answer 19

Oh, okay. Thank you very much for the help. Crystal clear.

Answer 20

youre welcome